(a) To find the coordinates of \(A\), solve the system of equations given by the circle \(x^2 + y^2 = 2\) and the line \(y = 2x - 1\). Substitute \(y = 2x - 1\) into the circle equation:

\(x^2 + (2x - 1)^2 = 2\)

\(x^2 + 4x^2 - 4x + 1 = 2\)

\(5x^2 - 4x - 1 = 0\)

Solving this quadratic equation, we find \(x = 1\). Substituting back, \(y = 2(1) - 1 = 1\). Thus, \(A = (1, 1)\).

(b) The volume of revolution is given by \(\pi \int (2 - x^2) \, dx\) from \(x = 1\) to \(x = \sqrt{2}\). The integral is:

\(\pi \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt{2}}\)

\(\pi \left[ 2\sqrt{2} - \frac{(\sqrt{2})^3}{3} - (2 - \frac{1}{3}) \right]\)

\(\pi \left[ 2\sqrt{2} - \frac{2\sqrt{2}}{3} - \frac{5}{3} \right]\)

\(\frac{\pi}{3}(4\sqrt{2} - 5)\)

(c) The perimeter of the shaded region consists of the line segment \(AD\) and the arc length. The arc length is \(\frac{1}{8}(2\pi\sqrt{2})\) or \(\frac{\pi\sqrt{2}}{4}\). Therefore, the perimeter is:

\(\sqrt{2} + \frac{\pi\sqrt{2}}{4}\)